-0.000 282 005 938 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 938(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 282 005 938(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 282 005 938| = 0.000 282 005 938


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 282 005 938.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 282 005 938 × 2 = 0 + 0.000 564 011 876;
  • 2) 0.000 564 011 876 × 2 = 0 + 0.001 128 023 752;
  • 3) 0.001 128 023 752 × 2 = 0 + 0.002 256 047 504;
  • 4) 0.002 256 047 504 × 2 = 0 + 0.004 512 095 008;
  • 5) 0.004 512 095 008 × 2 = 0 + 0.009 024 190 016;
  • 6) 0.009 024 190 016 × 2 = 0 + 0.018 048 380 032;
  • 7) 0.018 048 380 032 × 2 = 0 + 0.036 096 760 064;
  • 8) 0.036 096 760 064 × 2 = 0 + 0.072 193 520 128;
  • 9) 0.072 193 520 128 × 2 = 0 + 0.144 387 040 256;
  • 10) 0.144 387 040 256 × 2 = 0 + 0.288 774 080 512;
  • 11) 0.288 774 080 512 × 2 = 0 + 0.577 548 161 024;
  • 12) 0.577 548 161 024 × 2 = 1 + 0.155 096 322 048;
  • 13) 0.155 096 322 048 × 2 = 0 + 0.310 192 644 096;
  • 14) 0.310 192 644 096 × 2 = 0 + 0.620 385 288 192;
  • 15) 0.620 385 288 192 × 2 = 1 + 0.240 770 576 384;
  • 16) 0.240 770 576 384 × 2 = 0 + 0.481 541 152 768;
  • 17) 0.481 541 152 768 × 2 = 0 + 0.963 082 305 536;
  • 18) 0.963 082 305 536 × 2 = 1 + 0.926 164 611 072;
  • 19) 0.926 164 611 072 × 2 = 1 + 0.852 329 222 144;
  • 20) 0.852 329 222 144 × 2 = 1 + 0.704 658 444 288;
  • 21) 0.704 658 444 288 × 2 = 1 + 0.409 316 888 576;
  • 22) 0.409 316 888 576 × 2 = 0 + 0.818 633 777 152;
  • 23) 0.818 633 777 152 × 2 = 1 + 0.637 267 554 304;
  • 24) 0.637 267 554 304 × 2 = 1 + 0.274 535 108 608;
  • 25) 0.274 535 108 608 × 2 = 0 + 0.549 070 217 216;
  • 26) 0.549 070 217 216 × 2 = 1 + 0.098 140 434 432;
  • 27) 0.098 140 434 432 × 2 = 0 + 0.196 280 868 864;
  • 28) 0.196 280 868 864 × 2 = 0 + 0.392 561 737 728;
  • 29) 0.392 561 737 728 × 2 = 0 + 0.785 123 475 456;
  • 30) 0.785 123 475 456 × 2 = 1 + 0.570 246 950 912;
  • 31) 0.570 246 950 912 × 2 = 1 + 0.140 493 901 824;
  • 32) 0.140 493 901 824 × 2 = 0 + 0.280 987 803 648;
  • 33) 0.280 987 803 648 × 2 = 0 + 0.561 975 607 296;
  • 34) 0.561 975 607 296 × 2 = 1 + 0.123 951 214 592;
  • 35) 0.123 951 214 592 × 2 = 0 + 0.247 902 429 184;
  • 36) 0.247 902 429 184 × 2 = 0 + 0.495 804 858 368;
  • 37) 0.495 804 858 368 × 2 = 0 + 0.991 609 716 736;
  • 38) 0.991 609 716 736 × 2 = 1 + 0.983 219 433 472;
  • 39) 0.983 219 433 472 × 2 = 1 + 0.966 438 866 944;
  • 40) 0.966 438 866 944 × 2 = 1 + 0.932 877 733 888;
  • 41) 0.932 877 733 888 × 2 = 1 + 0.865 755 467 776;
  • 42) 0.865 755 467 776 × 2 = 1 + 0.731 510 935 552;
  • 43) 0.731 510 935 552 × 2 = 1 + 0.463 021 871 104;
  • 44) 0.463 021 871 104 × 2 = 0 + 0.926 043 742 208;
  • 45) 0.926 043 742 208 × 2 = 1 + 0.852 087 484 416;
  • 46) 0.852 087 484 416 × 2 = 1 + 0.704 174 968 832;
  • 47) 0.704 174 968 832 × 2 = 1 + 0.408 349 937 664;
  • 48) 0.408 349 937 664 × 2 = 0 + 0.816 699 875 328;
  • 49) 0.816 699 875 328 × 2 = 1 + 0.633 399 750 656;
  • 50) 0.633 399 750 656 × 2 = 1 + 0.266 799 501 312;
  • 51) 0.266 799 501 312 × 2 = 0 + 0.533 599 002 624;
  • 52) 0.533 599 002 624 × 2 = 1 + 0.067 198 005 248;
  • 53) 0.067 198 005 248 × 2 = 0 + 0.134 396 010 496;
  • 54) 0.134 396 010 496 × 2 = 0 + 0.268 792 020 992;
  • 55) 0.268 792 020 992 × 2 = 0 + 0.537 584 041 984;
  • 56) 0.537 584 041 984 × 2 = 1 + 0.075 168 083 968;
  • 57) 0.075 168 083 968 × 2 = 0 + 0.150 336 167 936;
  • 58) 0.150 336 167 936 × 2 = 0 + 0.300 672 335 872;
  • 59) 0.300 672 335 872 × 2 = 0 + 0.601 344 671 744;
  • 60) 0.601 344 671 744 × 2 = 1 + 0.202 689 343 488;
  • 61) 0.202 689 343 488 × 2 = 0 + 0.405 378 686 976;
  • 62) 0.405 378 686 976 × 2 = 0 + 0.810 757 373 952;
  • 63) 0.810 757 373 952 × 2 = 1 + 0.621 514 747 904;
  • 64) 0.621 514 747 904 × 2 = 1 + 0.243 029 495 808;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 282 005 938(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011(2)

6. Positive number before normalization:

0.000 282 005 938(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 938(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011(2) × 2-12


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1011(10) =

011 1111 0011(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011 =

0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011


Decimal number -0.000 282 005 938 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0100 0111 1110 1110 1101 0001 0001 0011

Share

» Convert decimal -0.000 282 005 916 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100